Can two admissable heuristics not dominate each other?
This is possible. Admissibility only asserts that the heuristic will never overestimate the true cost. With that being said, it is possible for one heuristic in some cases to do better than another and vice-versa. Think of it as a game of rock paper scissors.

Specifically, you may find that sometimes $h_1 &lt; h_2$ and in other times $h_2 &lt; h_1$, where $h_1$ and $h_2$ are admissible heuristics. Thus, by definition, neither strictly dominates the other. 

In fact, there is a way to "combine" the two admissible heuristics to get the best of both using:

$$h_3 = \max(h_1, h_2)$$